Question

A two-digit number exceeds by $19$ the sum of the squares of its digits and by $44$ the double product of its digits. Find the number.

Answer 1

Let $n(a,b)$ be two digit nubmer

$n(a,b)=10a+b$

Given $n(a,b)=a^2+b^2+18\to \left(1\right)$

$10a+b=a^2+b^2+19$

$n(a,b)=2ab+44\to \left(2\right)$

$\left(1\right)-\left(2\right)\Rightarrow (a-b{)}^2=25$

$a-b=\pm 5$

If $a=b-5$

$\left(2\right)\Rightarrow 11b-50=b^2-10b+25+b^2+19$

$2b^2-21b+94=0\to$ No solution

If $a=b+5$,

$\left(1\right)\Rightarrow 11b+50=b^2+10b+25+b^2+19$

$2b^2-b-6=0$

$(b-2)(2b+3)=0$

$\therefore b=2,a=7$

$\therefore n(a,b)=72$ is the required number.